你如何从给定的N个数字中测试所有可能的加法组合,使它们加起来得到给定的最终数字?

一个简单的例子:

要添加的数字集:N ={1,5,22,15,0,…} 期望结果:12345


当前回答

@KeithBeller的回答略有变化的变量名称和一些评论。

    public static void Main(string[] args)
    {
        List<int> input = new List<int>() { 3, 9, 8, 4, 5, 7, 10 };
        int targetSum = 15;
        SumUp(input, targetSum);
    }

    public static void SumUp(List<int> input, int targetSum)
    {
        SumUpRecursive(input, targetSum, new List<int>());
    }

    private static void SumUpRecursive(List<int> remaining, int targetSum, List<int> listToSum)
    {
        // Sum up partial
        int sum = 0;
        foreach (int x in listToSum)
            sum += x;

        //Check sum matched
        if (sum == targetSum)
            Console.WriteLine("sum(" + string.Join(",", listToSum.ToArray()) + ")=" + targetSum);

        //Check sum passed
        if (sum >= targetSum)
            return;

        //Iterate each input character
        for (int i = 0; i < remaining.Count; i++)
        {
            //Build list of remaining items to iterate
            List<int> newRemaining = new List<int>();
            for (int j = i + 1; j < remaining.Count; j++)
                newRemaining.Add(remaining[j]);

            //Update partial list
            List<int> newListToSum = new List<int>(listToSum);
            int currentItem = remaining[i];
            newListToSum.Add(currentItem);
            SumUpRecursive(newRemaining, targetSum, newListToSum);
        }
    }'

其他回答

c#版本的@msalvadores代码的答案

void Main()
{
    int[] numbers = {3,9,8,4,5,7,10};
    int target = 15;
    sum_up(new List<int>(numbers.ToList()),target);
}

static void sum_up_recursive(List<int> numbers, int target, List<int> part)
{
   int s = 0;
   foreach (int x in part)
   {
       s += x;
   }
   if (s == target)
   {
        Console.WriteLine("sum(" + string.Join(",", part.Select(n => n.ToString()).ToArray()) + ")=" + target);
   }
   if (s >= target)
   {
        return;
   }
   for (int i = 0;i < numbers.Count;i++)
   {
         var remaining = new List<int>();
         int n = numbers[i];
         for (int j = i + 1; j < numbers.Count;j++)
         {
             remaining.Add(numbers[j]);
         }
         var part_rec = new List<int>(part);
         part_rec.Add(n);
         sum_up_recursive(remaining,target,part_rec);
   }
}
static void sum_up(List<int> numbers, int target)
{
    sum_up_recursive(numbers,target,new List<int>());
}

我不喜欢上面看到的Javascript解决方案。下面是我使用部分应用、闭包和递归构建的一个:

好的,我主要关心的是,如果组合数组能满足目标要求,希望这样你就能找到剩下的组合了

这里只需要设置目标并传递组合数组。

function main() {
    const target = 10
    const getPermutationThatSumT = setTarget(target)
    const permutation = getPermutationThatSumT([1, 4, 2, 5, 6, 7])

    console.log( permutation );
}

我提出的当前实现

function setTarget(target) {
    let partial = [];

    return function permute(input) {
        let i, removed;
        for (i = 0; i < input.length; i++) {
            removed = input.splice(i, 1)[0];
            partial.push(removed);

            const sum = partial.reduce((a, b) => a + b)
            if (sum === target) return partial.slice()
            if (sum < target) permute(input)

            input.splice(i, 0, removed);
            partial.pop();
        }
        return null
    };
}
Thank you.. ephemient

我已经将上述逻辑从python转换为php..

<?php
$data = array(array(2,3,5,10,15),array(4,6,23,15,12),array(23,34,12,1,5));
$maxsum = 25;

print_r(bestsum($data,$maxsum));  //function call

function bestsum($data,$maxsum)
{
$res = array_fill(0, $maxsum + 1, '0');
$res[0] = array();              //base case
foreach($data as $group)
{
 $new_res = $res;               //copy res

  foreach($group as $ele)
  {
    for($i=0;$i<($maxsum-$ele+1);$i++)
    {   
        if($res[$i] != 0)
        {
            $ele_index = $i+$ele;
            $new_res[$ele_index] = $res[$i];
            $new_res[$ele_index][] = $ele;
        }
    }
  }

  $res = $new_res;
}

 for($i=$maxsum;$i>0;$i--)
  {
    if($res[$i]!=0)
    {
        return $res[$i];
        break;
    }
  }
return array();
}
?>

PHP版本,灵感来自Keith Beller的c#版本。

bala的PHP版本不适合我,因为我不需要对数字进行分组。我想要一个更简单的实现,只有一个目标值和一个数字池。这个函数也会删除任何重复的条目。

编辑25/10/2021:添加精度参数以支持浮点数(现在需要bcmath扩展)。

/**
 * Calculates a subset sum: finds out which combinations of numbers
 * from the numbers array can be added together to come to the target
 * number.
 *
 * Returns an indexed array with arrays of number combinations.
 *
 * Example:
 *
 * <pre>
 * $matches = subset_sum(array(5,10,7,3,20), 25);
 * </pre>
 *
 * Returns:
 *
 * <pre>
 * Array
 * (
 *   [0] => Array
 *   (
 *       [0] => 3
 *       [1] => 5
 *       [2] => 7
 *       [3] => 10
 *   )
 *   [1] => Array
 *   (
 *       [0] => 5
 *       [1] => 20
 *   )
 * )
 * </pre>
 *
 * @param number[] $numbers
 * @param number $target
 * @param array $part
 * @param int $precision
 * @return array[number[]]
 */
function subset_sum($numbers, $target, $precision=0, $part=null)
{
    // we assume that an empty $part variable means this
    // is the top level call.
    $toplevel = false;
    if($part === null) {
        $toplevel = true;
        $part = array();
    }

    $s = 0;
    foreach($part as $x)
    {
        $s = $s + $x;
    }

    // we have found a match!
    if(bccomp((string) $s, (string) $target, $precision) === 0)
    {
        sort($part); // ensure the numbers are always sorted
        return array(implode('|', $part));
    }

    // gone too far, break off
    if($s >= $target)
    {
        return null;
    }

    $matches = array();
    $totalNumbers = count($numbers);

    for($i=0; $i < $totalNumbers; $i++)
    {
        $remaining = array();
        $n = $numbers[$i];

        for($j = $i+1; $j < $totalNumbers; $j++)
        {
            $remaining[] = $numbers[$j];
        }

        $part_rec = $part;
        $part_rec[] = $n;

        $result = subset_sum($remaining, $target, $precision, $part_rec);
        if($result)
        {
            $matches = array_merge($matches, $result);
        }
    }

    if(!$toplevel)
    {
        return $matches;
    }

    // this is the top level function call: we have to
    // prepare the final result value by stripping any
    // duplicate results.
    $matches = array_unique($matches);
    $result = array();
    foreach($matches as $entry)
    {
        $result[] = explode('|', $entry);
    }

    return $result;
}

例子:

$result = subset_sum(array(5, 10, 7, 3, 20), 25);

这将返回一个包含两个数字组合数组的索引数组:

3, 5, 7, 10
5, 20

浮点数示例:

// Specify the precision in the third argument
$result = subset_sum(array(0.40, 0.03, 0.05), 0.45, 2);

这将返回一个匹配项:

0.40, 0.05

这是R中的一个解

subset_sum = function(numbers,target,partial=0){
  if(any(is.na(partial))) return()
  s = sum(partial)
  if(s == target) print(sprintf("sum(%s)=%s",paste(partial[-1],collapse="+"),target))
  if(s > target) return()
  for( i in seq_along(numbers)){
    n = numbers[i]
    remaining = numbers[(i+1):length(numbers)]
    subset_sum(remaining,target,c(partial,n))
  }
}